paix1988 said:
*I am struggling with arithmetic and geometric sequences.
if the 4th term is m-8, 6th term 8m+3 and 8th term is 10m-5
Calculate the 1st and 5th term
Which term will have a value of -70The 4th term of geometric sequence is -16 and the 6th term is -64. Calculate the 3rd and 5th terms.
thank you for your assistance
For the benefit of the community, I am going to post solutions.
1.) We are given the following information regarding an arithmetic progression:
$$a_4=m-8,\,a_6=8m+3,\,a_8=10m-5$$
If we denote the common difference as $d$, then from this we know:
$$2d=(8m+3)-(m-8)=7m+11$$
$$2d=(10m-5)-(8m+3)=2m-8$$
Equating the two results, we obtain:
$$7m+11=2m-8\implies m=-\frac{19}{5}\implies d=-\frac{39}{5}$$
Now, in general, we have:
$$a_n=a_1+(n-1)d$$
And we know:
$$a_4=a_1+3d$$
$$m-8=a_1+3d$$
$$-\frac{59}{5}=a_1-\frac{117}{5}$$
$$a_1=\frac{58}{5}$$
Hence, the general term is:
$$a_n=\frac{58}{5}+(n-1)\left(-\frac{39}{5}\right)=\frac{58-39(n-1)}{5}$$
Thus, the 5th term is:
$$a_5=\frac{58-39(5-1)}{5}=-\frac{98}{5}$$
To find which term has a value of -70, we may write:
$$\frac{58-39(n-1)}{5}=-70$$
$$58-39(n-1)=-350$$
$$-39(n-1)=-408$$
$$n-1=\frac{136}{13}$$
We will not get an integral value for $n$, thus none of the terms of this AP has a value of -70.
2.)The $n$th term of a GP is:
$$a_n=ar^n$$
We know that:
$$\frac{g_6}{g_4}=\frac{-64}{-16}=4=r^2\implies r=\pm2\implies a=-1$$
Thus, there are two possible GPs satisfying the given information:
a) $$a_n=-(-2)^n\implies a_3=8,\,a_5=32$$
b) $$a_n=-2^n\implies a_3=-8,\,a_5=-32$$